Answer to Question #198209 in Molecular Physics | Thermodynamics for Charles

Question

Two men are standing on the same side of a high wall and at the same distance from it such that they are 400 m apart. when one man fires a pistol, the other hears the first report 1.2 s after seeing the flash and the second report 0.5 s after the first. Explain why he hears two reports and calculate the (a) velocity of sound in air (b) perpendicular distance of the men from the wall?

Answer 1

Answer: we will solve like this

First we assume that two men are A and B

Distance between A and B= 400 m

time for first sound that B report= 1.2 s

time for second sound (echo) that B report= 0.5s

Velocity of sound in air= distance/time$= \frac{400}{1.2}$ = 333.33 ms⁻¹

Echo time= 1.2+0.5= 1.7s

Echo distance = Speed of sound in air $\times$ Echo time

             = 333.33×1.7= 566.661 m

As two sides of the triangle are congruent, one congruent side will be=

             = 566.661/2= 283.35 m

In a isosceles right angle triangle two of the angles will be of 45°, as in ΔCBE; therefore perpendicular distance BE will be$= sin45°\times BC$ (one congruent side caluclated in ΔABC)= 0.707×283.335

                 Ans  =200 m perpendicular distance of the men from the wall